Orthogonality, Projections and Linear Systems Previous Year Questions (PYQs) for GATE DA: 2+ Solved Questions with Step-by-Step Solutions

    Solve 2+ Orthogonality, Projections and Linear Systems previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

    Chapter Roadmap: Orthogonality, Projections and Linear Systems

    Chapter Journey

    Step 1: Projection Matrices, Null Space and Idempotence (Current)
    Geometric intuition of orthogonal projections. Algebraic properties: Idempotence and Symmetry. Column space, Null space, and Eigenvalues.
    Step 2: Consistency and Solution Sets of Linear Systems
    Conditions for a linear system to have a solution. Geometric interpretation of consistency. Structuring the complete solution set.
    Goal: Master the transition from projecting vectors to solving and analyzing linear systems.

    The Geometry of Projection Matrices

    U b p e

    The Core Idea

    A projection matrix maps any vector to a vector in a target subspace , such that the error vector is orthogonal to .

    Geometric Meaning
    • is the "shadow" or closest point in .
    • is the "error" or perpendicular drop.
    • isolates the component of that lives inside .

    Orthogonality, Projections and Linear Systems: Solved Questions with Step-by-Step Explanations (2 Problems)

    Question 1 · Linear Algebra MSQ
    Let be the set of real numbers, be a subspace of and be the
    matrix corresponding to the projection on to the subspace .
    Which of the following statements is/are TRUE?
    1. A. If is a 1-dimensional subspace of , then the null space of is a
      1-dimensional subspace.
    2. B. If is a 2-dimensional subspace of , then the null space of is a
      1-dimensional subspace.
    3. C.

    4. D.

    Correct Answer:

    ["B","C","D"]

    Step-by-Step Solution

    Key idea: This question tests the fundamental algebraic and geometric properties of orthogonal projection matrices, specifically the Rank-Nullity Theorem and idempotence.

    Step 1: Analyze the Null Space dimension (Options A and B).

    Let be the projection matrix onto subspace .

    The column space of , denoted , is exactly the target subspace . Thus, .

    By the Rank-Nullity Theorem for a matrix:

    Check Option A: If , then . Option A claims it is 1-dimensional. This is FALSE.

    Check Option B: If , then . Option B claims it is 1-dimensional. This is TRUE.

    Step 2: Analyze Idempotence (Option C).

    A matrix is a projection matrix if and only if it is idempotent, meaning .

    Geometrically, if you project a vector onto to get , projecting again yields itself because is already in .

    Algebraically: .

    Thus, Option C is TRUE.

    Step 3: Analyze Higher Powers (Option D).

    Since , we can compute :

    By induction, for all integers .

    Thus, is TRUE.

    Answer: Options B, C, and D are TRUE.

    Question 2 · Linear Algebra MSQ
    Which of the following statements is/are TRUE?
    Note: denotes the set of real numbers.
    1. A. There exist , , and such that has a unique
      solution and has infinite solutions.
    2. B. There exist , , and such that has no solutions
      and has infinite solutions.
    3. C. There exist , , and such that has a unique
      solution and has infinite solutions.
    4. D. There exist , , and such that has a unique
      solution and has no solutions.
    Correct Answer:

    ["B","D"]

    Step-by-Step Solution

    Key idea: This question tests the consistency conditions of linear systems based on the shape of matrix (square, wide, or tall) and its rank.

    Step 1: Analyze Option A ().

    For a square matrix , if has a unique solution, then ( is invertible).

    If is invertible, must also have a unique solution () for ANY .

    It is impossible for an invertible matrix to have infinite solutions for any RHS.

    Thus, Option A is FALSE.

    Step 2: Analyze Option B ().

    Can have no solution and have infinite solutions?

    Yes. Let be singular (Rank < 3).

    Example: .

    • Let . System is inconsistent (no solution) because .
    • Let . System is . is free. Infinite solutions.

    Thus, Option B is TRUE.

    Step 3: Analyze Option C ().

    Can have a unique solution?

    has 3 columns (variables) and 2 rows (equations).

    Max Rank is 2.

    Number of free variables = .

    If a solution exists, there is at least 1 free variable, implying infinite solutions.

    A unique solution is impossible for a wide matrix ().

    Thus, Option C is FALSE.

    Step 4: Analyze Option D ().

    Can have a unique solution?

    has 2 columns. Max Rank is 2.

    If Rank()=2, the columns are linearly independent.

    If is in the column space, the solution is unique (0 free variables).

    Can have no solution?

    Yes, if is not in the column space (which is a 2D plane in ).

    So, Unique Solution AND No Solution are both possible for different RHS vectors.

    Thus, Option D is TRUE.

    Answer: Options B and D are TRUE.

    More previous year questions (pyqs) in this unit

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    Orthogonality, Projections and Linear Systems Previous Year Questions (PYQs) for GATE DA: 2+ Solved Questions with Step-by-Step Solutions

    Solve 2+ Orthogonality, Projections and Linear Systems previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

    A question from this chapter

    Question 1
    Let be the set of real numbers, be a subspace of and be the
    matrix corresponding to the projection on to the subspace .
    Which of the following statements is/are TRUE?
    Question 2
    Which of the following statements is/are TRUE?
    Note: denotes the set of real numbers.
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